Let $g(x)=6 x^4-32 x^3+48 x^2-7$. What is the absolute minimum value of $g$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-64$ (Choice B) B $-25$ (Choice C) C $-7$ (Choice D) D $g$ has no minimum value
Let's first find the relative extremum points of $g$, and then consider them along with the function's end behavior in both directions. We start with finding the critical points of $g$. The derivative of $g$ is $g'(x)=24x(x-2)^2$. $g'(x)=0$ for $x=0,2$. $g'$ is defined for all real numbers. Therefore, our critical points are $x=0$ and $x=2$. Our critical points divide the function's domain (which is all real numbers) into three intervals: $\llap{-}1$ $0$ $1$ $2$ $3$ $(-\infty,0)$ $(0,2)$ $(2,\infty)$ Let's evaluate $g'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g'(x)$ Verdict $(-\infty,0)$ $x=-1$ $g'(-1)=-216<0$ $g$ is decreasing $\searrow$ $(0,2)$ $x=1$ $g'(1)=24>0$ $g$ is increasing $\nearrow$ $(2,\infty)$ $x=3$ $g'(3)=72>0$ $g$ is increasing $\nearrow$ Now let's look at all the critical points: $x$ $g(x)$ Before After Verdict $0$ $-7$ $\searrow$ $\nearrow$ Minimum $2$ $25$ $\nearrow$ $\nearrow$ Not an extremum Let's imagine ourselves walking on the graph of $g$, starting all the way to the left (from $-\infty$ ) and going all the way to the right (until $+\infty$ ). According to the table, we will start by going down until $(0,-7)$, and then forever go up. This means that $\lim_{x\to-\infty}g(x)=\lim_{x\to +\infty}g(x)=+\infty$, which means $g$ has no maximum value. However, $g$ does reach an absolute minimum point at $(0,-7)$, which means its absolute minimum value is $-7$. In conclusion, the absolute minimum value of $g$ is $-7$.